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# A number N when expressed as product of prime factors gives.

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A number N when expressed as product of prime factors gives. [#permalink]

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06 Oct 2010, 05:07
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A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027
B.3125
C.243
D.729
E.None of these

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Re: A number N when expressed as product of prime factors gives. [#permalink]

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06 Oct 2010, 14:25
pzazz12 wrote:
A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A. 1027
B. 3125
C. 243
D. 729
E. None of these

Question is quite ambiguous but here is what I think:

So we are looking for all cases of N = x * y * z like:

N = (1) * (1) * (2^5 * 3^2 * 5^4 * 7 * 11^3);
N = (1) * (2^5) * (3^2 * 5^4 * 7 * 11^3);
N = (1) * (2^5 * 3^2) * (5^4 * 7 * 11^3);
N = (2^5) * (3^2) * (5^4 * 7 * 11^3);
N = (2^5 * 3^2) * (5^4 * 7) * (11^3);
...

In this case factors x * y * z would be co-prime (wont share any common factor but 1) as each prime will be only in one factor.

As there are total of 5 primes in N then there are following cases possible:

1. {1}*{1}*{factor with all five primes} - 1 (1*1*N);

2. {1}*{factor with one prime}*{factor with four primes} - $$C^1_5*C^4_4=5$$ ($$C^1_5$$ - choosing which prime will be in one-prime factor, the rest primes go to the third factor);

3. {1}*{factor with two prime}*{factor with three primes} - $$C^2_5*C^3_3=10$$ ($$C^2_5$$ - choosing which 2 primes will be in two-prime factor, the rest primes go to the third factor);

4. {factor with one prime}*{factor with one prime}*{factor with three primes} - $$C^3_5=10$$ ($$C^3_5$$ - choosing which 3 primes will be in three-prime factor, from the 2 primes left one will go to the first factor and another to the second);

5. {factor with one prime}*{factor with two primes}*{factor with two primes} - $$C^1_5*\frac{C^2_4}{2}=15$$ ($$C^1_5$$ - choosing which 1 primes will be in one-prime factor, 4 primes left can be split among two factors (into two groups) in $$\frac{C^2_4}{2}$$ ways, dividing by 2 as the order of the factors (groups) does not matter);

Total: $$1+5+10+10+15=41$$.

No such answer among answer choices.

So I guess it's meant that the order of the factors is important, for example N= (2^5) * (3^2) * (5^4 * 7 * 11^3) is different from N= (3^2) * (2^5) * (5^4 * 7 * 11^3), then case 1 can be arranged in 3 ways and cases 2, 3, 4, and 5 in 3! ways: $$3*1+3!(5+10+10+15)=243$$.

So answer: C.

But if the order of the factors is important then the problem can be solved easier: each of the five primes can be part of the first, the second or the third factor so each prime has 3 options. Total ways to distribute 5 primes would be $$3*3*3*3*3=3^5=243$$.

Answer: C.

Anyway not a GMAT question, so I wouldn't worry about it at all.
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Re: A number N when expressed as product of prime factors gives. [#permalink]

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06 Oct 2010, 14:42
pzazz12 wrote:
A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027
B.3125
C.243
D.729
E.None of these

The trick is treat $$2^5, 3^2, 5^4, 7, 11^3$$ as 5 objects which you need to place in 3 groups. Order being irrelevant as that cannot matter

So the answer would be :
$$\frac{5!}{0!0!5!} + \frac{5!}{1!0!4!} + \frac{5!}{1!1!3!} + \frac{5!}{2!0!3!} + \frac{5!}{2!1!2!}$$
$$= 1 + 5 + 20 + 10 + 30 =66$$

Now its been too long since I learnt P&C, and this is a hard question ... so I am not 100% confident, but I would guess (e)

What I am 100% confident about is that this is beyond a GMAT difficultly level
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Re: A number N when expressed as product of prime factors gives. [#permalink]

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06 Oct 2010, 15:11
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shrouded1 wrote:
pzazz12 wrote:
A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027
B.3125
C.243
D.729
E.None of these

The trick is treat $$2^5, 3^2, 5^4, 7, 11^3$$ as 5 objects which you need to place in 3 groups. Order being irrelevant as that cannot matter

So the answer would be :
$$\frac{5!}{0!0!5!} + \frac{5!}{1!0!4!} + \frac{5!}{1!1!3!} + \frac{5!}{2!0!3!} + \frac{5!}{2!1!2!}$$
$$= 1 + 5 + 20 + 10 + 30 =66$$

Now its been too long since I learnt P&C, and this is a hard question ... so I am not 100% confident, but I would guess (e)

What I am 100% confident about is that this is beyond a GMAT difficultly level

I think 66 contains duplication.

For example if we manually count case 1-1-3 we will get:
{2}-{3}-{5, 7, 11}
{2}-{5}-{3, 7, 11}
{2}-{7}-{3, 5, 11}
{2}-{11}-{3, 5, 7}
{5}-{3}-{2, 7, 11}
{7}-{3}-{2, 5, 11}
{11}-{3}-{2, 5, 7}
{5}-{7}-{2, 3, 11}
{5}-{11}-{2, 3, 7}
{7}-{11}-{2, 3, 5}

Total of 10 cases (when order is not important) but as per your formula it's $$\frac{5!}{1!1!3!}=20$$. The same with 1-2-2, it's also twice as much. If you subtract this duplications 10 and 15 then you'll get 41 as in my calculations.
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Re: A number N when expressed as product of prime factors gives. [#permalink]

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06 Oct 2010, 15:17
Good spot ! Agreed.

I can intuitively tell why the duplication exists. Its sort of because the two "equal" groups in 3-1-1 and 2-2-1 are somehow treated as "distint" in these formulae. I agree with the 41. Surely order can't metter though
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Re: A number N when expressed as product of prime factors gives. [#permalink]

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08 Oct 2010, 03:29
I've seen this question being posted on beatthegmat.com and according to experts there, it is beyond GMAT. How can one solve this question within 2 minutes? or even 3 mins I think. "pairwise co-prime" statement is ambiguous.
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Re: A number N when expressed as product of prime factors gives. [#permalink]

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24 Aug 2012, 08:04
Irrelevant practice for the GMAT but I am sure it's fun for those who solved it.
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Re: A number N when expressed as product of prime factors gives. [#permalink]

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22 Nov 2016, 08:46
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pzazz12 wrote:
A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027
B.3125
C.243
D.729
E.None of these

“Pairwise coprime” means that if we choose any 2 numbers from a given set of numbers they will be co-prime to each other.

This question is about splitting into groups*.

We have 5 possible cases including those which have 1 as a factor:

1*1*(group of 5 primes)
1*(group of 1 prime)(group of 4 primes)
1*(group of 2 primes) (group of 3 primes)
(group of 1 prime)*(group of 1 prime)*(group of 3 primes)
(group of 1prime)*(group of 2 primes)*(group of 2 primes)

* - in each group primes are different and choosing 1 as a factor means choosing 0 from a group of a given 5 different prime factors.

Values of given answer options suggest that order of factors matters (we won’t be able to achieve such big numbers otherwise).

Hence, we have:

$$\frac{5!}{0!*0!*5!} * \frac{3!}{2!} + \frac{5!}{0!*1!*4!} *3! + \frac{5!}{0!*2!*3!} *3! + \frac{5!}{1!*1!*3!} *\frac{3!}{2!} + \frac{5!}{1!*2!*2!} * \frac{3!}{2!} =$$

$$= 3 + 30 + 60 + 60 + 90 = 243$$

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Re: A number N when expressed as product of prime factors gives.   [#permalink] 22 Nov 2016, 08:46
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# A number N when expressed as product of prime factors gives.

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